Difference between revisions of "2004 AMC 10A Problems/Problem 18"
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Substituting <math>d=10</math> into the geometric progression gives the terms 9, 21, and 49; substituting <math>d=14</math> gives 9, -3, and 1. Therefore, the smallest possible 3rd term is 1 <math>\Rightarrow\mathrm{(A)}</math>. | Substituting <math>d=10</math> into the geometric progression gives the terms 9, 21, and 49; substituting <math>d=14</math> gives 9, -3, and 1. Therefore, the smallest possible 3rd term is 1 <math>\Rightarrow\mathrm{(A)}</math>. | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 02:41, 11 September 2007
Problem
A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the secon term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?
Solution
Let d be the difference between terms in the arithmetic progression, such that first three terms are 9, 9+d, and 9+2d. The terms of the geometric progression will be 9, 11+d, and 29+2d. Because they are in a geometric progression, we can say
, note that the terms of a geometric progression would be 
, x, and 
, so 
.
Solve this equation.
Substituting 
into the geometric progression gives the terms 9, 21, and 49; substituting 
gives 9, -3, and 1. Therefore, the smallest possible 3rd term is 1 
.
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
